Two bodies of masses m1 and m2 are connected by a light string which passes over a frictionless massless pulley . If the pulley is moving upward with uniform acceleration g/2 , then tension in the string is - 1) 3 m1 m2 g / m1 + m2 ..............2)m1 + m2 g / 4m1 m2 ..........3)2 m1 m2 g / m1 + m2 ..............4)m1 m2 g / m1 + m2 ..............
When the system is accelerating upward with acceleration (g/2), the net downward acceleration experienced by the masses is = g + g/2 = 3g/2
Now,
M(3g/2) – T = Ma ……..(1)
T – m(3g/2) = ma ………(2)
Adding the above two equations we get,
(3g/2)(M – m) = a(M + m)
=> a = (3g/2)(M – m)/(M + m)
So,
(2) => T = m(3g/2) + ma
=> T = m(3g/2) + (m)(3g/2)(M – m)/(M + m)
=> T = 3mgM/(M + m)
This is the tension in the string.
Correct option is (1).